14_InstSolManual_PDF_Part10 - 14-10 Chapter 14 14.57 Set Up...

14.57.Set Up:For water,Solve:The heat conducted by the rod in 10.0 min isReflect:The heat conducted by the rod is the heat that enters the ice and produces the phase change.14.58.Set Up:For copper,For steel,is the same for both sections ofthe rod.Solve: (a)For the copper section,(b)For the steel section,14.59.Set Up:Let the temperature of the fat-air boundary be T. A section of the two layers is sketched in Fig-ure 14.59. A Kelvin degree is the same size as a Celsius degree, so and are equivalent units. Atsteady state the heat current through each layer is the same, equal to 50 W. The area of each layer is withFigure 14.59Solve: (a)Apply to the fat layer and solve for For the fat layer (b)Apply to the air layer and solve for For the air layer and Reflect:The thermal conductivity of air is much lass than the thermal conductivity of fat, so the temperature gradi-ent for the air must be much larger to achieve the same heat current. So, most of the temperature difference is acrossthe air layer.L5kA

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